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# 1! + (2^2)! + 3! + (4^2)! +...+ 37! is divided by 5. What is

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Manager
Joined: 28 Aug 2004
Posts: 205

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1! + (2^2)! + 3! + (4^2)! +...+ 37! is divided by 5. What is [#permalink]

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04 Jan 2005, 08:59
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

1! + (2^2)! + 3! + (4^2)! +...+ 37! is divided by 5. What is the remainder?

A) 1
B) 2
C) 3
D) 4
E) 0

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VP
Joined: 18 Nov 2004
Posts: 1430

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04 Jan 2005, 09:44
"A". Will explain if correct.

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Manager
Joined: 29 Jul 2004
Posts: 61

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04 Jan 2005, 10:20
The pattern seems to be as follows:

sum of the numbers: 1 25 31

remiainder /5: 1 0 1

I stopped there because 16! would be too much to calculate. The 36th term would be (36^2)!, so the remainder would be 0 at this point based on the pattern, and the last term is 37!, so the remainder should be 1.

So my answer is A

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Senior Manager
Joined: 20 Dec 2004
Posts: 251

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04 Jan 2005, 17:00
The way I solved it: (probably similar to the previous post).

1! + (2^2)! + 3! + (4^2)! + ... + 37!

= 1 + 4! + 6 + 16! + ... + 37!

= 25 + 6 + 16! + ... + 37!

Based on the above expression 25 would result in a remainder of 0, 6 would result in a remainder of 1 and every term after that all the way to 37! will always result in a remainder of 0, as 5 will be a part of each of them (due to the factorial). Hence remainder is 1.

Kudos [?]: 112 [0], given: 0

Director
Joined: 31 Aug 2004
Posts: 606

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05 Jan 2005, 04:19
I considered that any number equal or greater than 5! does not provide any remainder so I just focused on the smallest terms.

1! + (2^2)! + 3! + (4^2)! +5!+(6^2)!...+ 37!

(4^2)!>5! and includes 5 so only 1!, 4! and 3! contribute to the remainder calculation.

=1+24+6=31

So remainder is 1.

Kudos [?]: 153 [0], given: 0

05 Jan 2005, 04:19
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# 1! + (2^2)! + 3! + (4^2)! +...+ 37! is divided by 5. What is

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